### NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

**Ex 16.2 Class 8 Maths Question 1.**

**If 21y5 is a multiple of 9, where y is a digit, what is the value of y?**

**Solution:**

A number is divisible by 9 if the sum of its digits is also divisible by 9.

Sum of the digits of 21y5 = 2 + 1 +y + 5 = 8 + y

(8 + y) ÷ 9 = 1

⇒ 8 + y = 9

⇒ y = 9 – 8 = 1

Hence, the required value of y = 1.

**Ex 16.2 Class 8 Maths Question 2.**

**If 31z5 is a multiple of 9, where z is a digit, what is the value of z?**

**Solution:**

A number is a multiple of 9 when the sum of its digits is also divisible by 9.

Sum of the digits of 31z5 = 3 + 1 + z + 5

3 + 1 + 2 + 5 = 9k where k is an integer.

For k = 1,

3 + 1 + z + 5 = 9

⇒ z = 9 – 9 = 0

For k = 2,

3 + 1 + z + 5 = 18

⇒ z = 18 – 9 = 9

k = 3 is not possible because 3 + 1 + z + 5 = 27

⇒ z = 27 – 9 = 18 which is not a digit.

Hence the required value of z is 0 or 9

**Ex 16.2 Class 8 Maths Question 3.**

**If 24x is a multiple of 3, where x is a digit, what is the value of x?**

**Solution:**

Since 24x is a multiple of 3, the sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers; 0, 3, 6, 12, 15, 18, ……..

6 + x = 3k where k is any integer.

For k = 0,

6 + x = 3 × 0

⇒ 6 + x = 0

x = -6. Not possible

For k = 1,

6 + x = 3 × 1

⇒ 6 + x = 3

⇒ x = 3 – 6 = -3. Not possible

For k = 2,

6 + x = 3 × 2

⇒ 6 + x = 6

⇒ x = 6 – 6 = 0

2 + 4 + 0 = 6 multiple of 3

For k = 3,

6 + x = 3 × 3

⇒ x = 9 – 6 = 3

2 + 4 + 3 = 9 multiple of 3

For k = 4,

6 + x = 3 × 4

⇒ 6 + x = 12

⇒ x = 12 – 6 = 6

2 + 4 + 6 = 12 which is multiple of 3

For k = 5,

6 + x = 3 × 5

⇒ x = 15 – 6 = 9

2 + 4 + 9 = 15 which is multiple of 3

For k = 6,

6 + x = 3 × 6

⇒ x = 18 – 6 = 12 not possible as x is digit

Hence the required values of x are 0, 3, 6 or 9.

**Ex 16.2 Class 8 Maths Question 4.**

**If 31z5 is a multiple of 3, where z is a digit, what might be the value of z?**

**Solution:**

A number is a multiple of 3 if the sum of its digits is divisible by 3.

3 + 1 + z + 5 = 3k where k is an integer

⇒ 9 + z = 3k

⇒ z = 3k – 9

Here, k = 0, 1, 2 is not possible as z is a digit of the number.

For k = 3,

z = 3 × 3 – 9 = 9 – 9 = 0

9 + 0 = 9 multiple of 3

For k = 4,

z = 3 × 4 – 9 = 12 – 9 = 3

9 + 3 = 12 multiple of 3

For k = 5,

z = 3 × 5 – 9 = 15 – 9 = 6

9 + 6 = 15 multiple of 3

For k = 6,

z = 3 × 6 – 9 = 18 – 9 = 9

9 + 9 = 18 multiple of 3

For k = 7,

z = 3 × 7 – 9 = 21 – 9 = 12 not possible as z is a digit

Hence, the required values of 2 are 0, 3, 6 and 9.